# Definte Integrals

1. Sep 23, 2007

Definite Integrals

I think I got these, but I left my text at work, so I was hoping someone could confirm that these answers are correct?

a.)
$$\int_{-1}^{1}\frac{dx}{1+x^2}$$
$$=\tan^{-1}(1)-\tan^{-1}(-1) =-\frac{\pi}{2}$$

and

b.)
$$\int_0^{\ln5}e^x(3-4e^x)$$
$$=\int_0^{\ln5}[3e^x-4e^{2x}]dx$$
$$=3e^x-2e^{2x}]_0^{\ln5}=49$$

Thanks,
Casey

Last edited: Sep 23, 2007
2. Sep 23, 2007

### Dick

No on both. You have a horrendous sign error on the first. On the second, it's just plain all wrong, even though the integration is correct. What gives?

3. Sep 23, 2007

No sleep.

The first one I am al effed up on...I thought arctan(1)=pi/4 and arctan(-1)=3pi/4 ?? my calculator is saying it is +pi/2 What does give?

part 2....if the integration is correct, I don't know maybe I am putting it in the calculator wrong......maybe I should not need a calculator for this..but I suck at powers of e....even though I know they are supposed to be easy.

4. Sep 23, 2007

3e^{ln5)=15.....4e^{2ln5} is where I think I am messing it up...
I do not remember how to evaluate.....

....4*ln25=100?

5. Sep 23, 2007

$$\int_0^{\ln5}e^x(3-4e^x)$$
$$=\int_0^{\ln5}[3e^x-4e^{2x}]dx$$
$$=3e^x-2e^{2x}]_0^{\ln5}=86$$

Maybe?

6. Sep 23, 2007

### Dick

arctan(-1)=-pi/4. exp(2*ln(5))=5^2=25. Etc, etc. I think you need to get some sleep.

7. Sep 23, 2007

Getting up at 3:30 am for work is not all it's cracked up to be. You are right though...my 24 hour streak is about over. In the AM I will need to look over that e to the ln crap.

Thanks

8. Sep 24, 2007

$$\int_0^{\ln5}e^x(3-4e^x)$$
$$=\int_0^{\ln5}[3e^x-4e^{2x}]dx$$
$$=3e^x-2e^{2x}]_0^{\ln5}=36$$
....Can someone tell me if this is correct, or if I should just kill myself now :/

Casey

9. Sep 24, 2007

### Gib Z

Closer, its actually -36. Get some sleep man, I remember once I stayed awake 30 hours, I walked home laughing at nothing...not a pretty sight for people walking by I'd imagine. Your not too far off insanity mate.